Formula: Thin circular ring Moment of inertia Mass Radius
$$\class{brown}{I} ~=~ \class{brown}{m} \, r^2$$
$$\class{brown}{I} ~=~ \class{brown}{m} \, r^2$$
$$\class{brown}{m} ~=~ \frac{\class{brown}{I}}{r^2}$$
$$r ~=~ \sqrt{\frac{\class{brown}{I}}{\class{brown}{m}}}$$
Moment of inertia
$$ \class{brown}{I} $$ Unit $$ \mathrm{kg} \, \mathrm{m}^2 $$
According to \( M ~=~ I \, \alpha \) (\(\alpha\): angular acceleration), the moment of inertia determines how hard it is to generate a torque \(M\) on the body. Moment of inertia \(I\) depends on the mass distribution and on the choice of the axis of rotation. Here we calculate the moment of inertia of a thin ring/strip whose axis of rotation passes through the center.
Mass
$$ \class{brown}{m} $$ Unit $$ \mathrm{kg} $$
Total mass of the thin ring.
Radius
$$ r $$ Unit $$ \mathrm{m} $$
Radius of the ring. This is the distance of the mass points of the ring from the axis of rotation.